Question: Simplify and expand the following expression: $ \dfrac{4r}{2r - 9}+\dfrac{4r}{4r + 9} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(2r - 9)(4r + 9)$ Multiply the first term by $\dfrac{4r + 9}{4r + 9}$ $ \begin{align*} \dfrac{4r}{2r - 9} \times \dfrac{4r + 9}{4r + 9} & = \dfrac{(4r)(4r + 9)}{(2r - 9)(4r + 9)} \\ & = \dfrac{16r^2 + 36r}{(2r - 9)(4r + 9)}\end{align*} $ Multiply the second term by $\dfrac{2r - 9}{2r - 9}$ $ \begin{align*} \dfrac{4r}{4r + 9} \times \dfrac{2r - 9}{2r - 9} & = \dfrac{(4r)(2r - 9)}{(4r + 9)(2r - 9)} \\ & = \dfrac{8r^2 - 36r}{(4r + 9)(2r - 9)}\end{align*} $ Now we have: $ = \dfrac{16r^2 + 36r}{(2r - 9)(4r + 9)} + \dfrac{8r^2 - 36r}{(4r + 9)(2r - 9)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{16r^2 + 36r + 8r^2 - 36r}{(2r - 9)(4r + 9)} $ $ = \dfrac{24r^2}{(2r - 9)(4r + 9)}$ Expand the denominator: $ = \dfrac{24r^2}{8r^2 - 18r - 81}$